Question

Finding intersection of two arrays can be done if you sort the 2 arrays and then do a linear step through to record the common elements.
This would take O(nlogn) + O(nlogn) + O(n)

Alternatively, you could compare each element in the first array with each element in the second array and you would get a O(n^2) run-time complexity.

How can I prove the first approach is better than the second?

Answer 1

First of all, O(nlogn) + O(nlogn) + O(n) doesn’t make much sense, since O(f) is a set, not a number.

What you mean is O(nlogn + nlogn + n), which can be shown to be equal to O(nlogn). Just look at what it means for a function f to belong to the set O(g):

f is an element of O(g) iff there exists c>0 and x0 such that for all x>x0:

|f(x)| <= c*|g(x)|

By setting f=nlogn and g=nlogn+nlogn+n, it follows that f is in O(g), and hence that O(nlogn) == O(nlogn + nlogn + n).